Zoran Skoda
twistsym

Low order Lie algebra type Hopf algebroid of nc phase space of Lie type

ϕ μ ν=δ μ ν+12𝒞 μ ν+112(𝒞 2) μ ν1720(𝒞 4) μ ν+O(C 6) \phi^\nu_\mu = \delta^\nu_\mu + \frac{1}{2}\mathcal{C}^\nu_\mu + \frac{1}{12}(\mathcal{C}^2)^\nu_\mu - \frac{1}{720}(\mathcal{C}^4)^\nu_\mu + O(C^6)

L μ ν=x μ νL_\mu^\nu = x_\mu\partial^\nu, 𝒞 ν μ=𝒞 νλ μ λ\mathcal{C}^\mu_\nu = \mathcal{C}^\mu_{\nu\lambda}\partial^\lambda.

Normal ordered form of 1\mathcal{F}^{-1} up to the third order

1=11+12x μ ν𝒞 ν μ+x μ12[ ν(𝒞 2) ν μ+(𝒞 2) ν μ ν]+x μx μ8 ν ν𝒞 ν μ𝒞 ν μ \mathcal{F}^{-1} = 1\otimes 1 + \frac{1}{2} x_\mu\partial^\nu\otimes \mathcal{C}^\mu_\nu + \frac{x_\mu}{12}\left[ \partial^\nu\otimes (\mathcal{C}^2)^\mu_\nu + (\mathcal{C}^2)^\mu_\nu\otimes\partial^\nu \right] + \frac{x_\mu x_{\mu'}}{8}\partial^\nu\partial^{\nu'}\otimes\mathcal{C}^\mu_\nu\mathcal{C}^{\mu'}_{\nu'}
x μ24 ν𝒞 ρ σ𝒞 ν ρ𝒞 σ μ+x μx μ24[ ν ν𝒞 ν μ(𝒞 2) ν μ𝒞 ν μ(𝒞 2) ν μ ν ν] -\frac{x_\mu}{24}\partial^\nu\mathcal{C}^\sigma_\rho\otimes\mathcal{C}^\rho_\nu\mathcal{C}^\mu_\sigma +\frac{x_\mu x_{\mu'}}{24}\left[\partial^\nu\partial^{\nu'}\otimes\mathcal{C}^\mu_\nu(\mathcal{C}^2)^{\mu'}_{\nu'} - \mathcal{C}^\mu_\nu(\mathcal{C}^2)^{\mu'}_{\nu'} \otimes\partial^\nu \partial^{\nu'}\right]

Up to the second order in CC-s, and with the undeformed exchange rules taken into account, we can write the symmetrized version, ln( S 1)ln(\mathcal{F}^{-1}_S), as

14(L μ ν𝒞 ν μ𝒞 ν μL μ ν)196(L μ ν(𝒞 2) ν μ+(𝒞 2) ν μL μ ν)+196(L μ ν𝒞 λ μ𝒞 ν λ+𝒞 ν λL μ ν𝒞 λ μ) \frac{1}{4}(L_\mu^\nu\otimes\mathcal{C}^\mu_\nu-\mathcal{C}^\mu_\nu\otimes L_\mu^\nu) -\frac{1}{96} (L_\mu^\nu\otimes(\mathcal{C}^2)^\mu_\nu + (\mathcal{C}^2)^\mu_\nu\otimes L_\mu^\nu) + \frac{1}{96} (L_\mu^\nu\mathcal{C}^\mu_\lambda\otimes \mathcal{C}^\lambda_\nu+\mathcal{C}^\lambda_\nu\otimes L_\mu^\nu\mathcal{C}^\mu_\lambda)

Note that for N=0,1,2,...N = 0,1,2,...

ν ν(𝒞 N) ν μ𝒞 ν μ= ν𝒞 ν μ ν(𝒞 N) ν μ \partial^\nu\partial^{\nu'}\otimes(\mathcal{C}^N)^\mu_\nu\mathcal{C}^{\mu'}_{\nu'} = -\partial^\nu \mathcal{C}^{\mu'}_{\nu'} \otimes \partial^{\nu'}(\mathcal{C}^N)^\mu_\nu

For the right polarized form R 1=exp( αx α)exp( λx^ λ)\mathcal{F}^{-1}_R = exp(-\partial^\alpha\otimes x_\alpha) exp(\partial^\lambda\otimes\hat{x}_\lambda), using BCH formula up to third order we get

ln( 1= α(x^ αx α)12 α β[x α,x^ β]+112 α β γ[x^ α+x α,[x β,x^ γ]], ln(\mathcal{F}^{-1} = \partial^\alpha\otimes(\hat{x}_\alpha-x_\alpha) - \frac{1}{2} \partial^\alpha\partial^\beta\otimes [x_\alpha,\hat{x}_\beta] +\frac{1}{12}\partial^\alpha\partial^\beta\partial^\gamma\otimes [\hat{x}_\alpha+x_\alpha,[x_\beta,\hat{x}_\gamma]],

where for the third term we used the fact that β γ\partial^\beta\partial^\gamma is symmetric in βγ\beta\leftrightarrow\gamma hence we renamed the indices.

Now, up to the third order,

[x α,x^ β]=[x α,x ρ(δ β ρ+12𝒞 β ρ+112(𝒞 2) β ρ)]=12C αβ ρ112C βα σ𝒞 σ ρ112𝒞 β σC σα ρ [x_\alpha,\hat{x}_\beta] = [x_\alpha,x_\rho(\delta^\rho_\beta + \frac{1}{2}\mathcal{C}^\rho_\beta + \frac{1}{12}(\mathcal{C}^2)^\rho_\beta)] = -\frac{1}{2} C^\rho_{\alpha\beta} - \frac{1}{12} C_{\beta\alpha}^\sigma \mathcal{C}^\rho_\sigma - \frac{1}{12}\mathcal{C}^\sigma_\beta C^\rho_{\sigma\alpha}

where the first two terms are antisymmetric and will drop out after contracting with α β\partial^\alpha\partial^\beta. Thus

12 α β[x α,x^ β]=124(𝒞 2x ρ λ) -\frac{1}{2}\partial^\alpha\partial^\beta\otimes [x_\alpha,\hat{x}_\beta] = -\frac{1}{24}(\mathcal{C}^2\otimes x_\rho\partial^\lambda)

Similarly,

112 α β γ[x^ α+x α,[x β,x^ γ]]=1288 α β γ[x τ𝒞 α τ,x ρ𝒞 γ σC βγ ρ] \frac{1}{12}\partial^\alpha\partial^\beta\partial^\gamma\otimes [\hat{x}_\alpha+x_\alpha,[x_\beta,\hat{x}_\gamma]] = \frac{1}{288}\partial^\alpha\partial^\beta\partial^\gamma\otimes [x_\tau\mathcal{C}^\tau_\alpha,x_\rho\mathcal{C}^\sigma_\gamma C^\rho_{\beta\gamma}]

as other terms drop out by the symmetry. The commutator gives two terms which are equal and of opposite sign, after renaming the indices and contraction with the tensor factor α β γ\partial^\alpha\partial^\beta\partial^\gamma:

1288 α β γx ρ(C ατ ρ𝒞 γ σC βσ τ𝒞 α τC γτ σC βσ ρ)=1288((𝒞 2) σ ρ γx ρ𝒞 γ σ(𝒞 2) τ ρ γx ρ𝒞 α τ)=0 \frac{1}{288}\partial^\alpha\partial^\beta\partial^\gamma\otimes x_\rho(C^\rho_{\alpha\tau}\mathcal{C}^\sigma_\gamma C^\tau_{\beta\sigma} -\mathcal{C}^\tau_\alpha C^\sigma_{\gamma\tau} C^\rho_{\beta\sigma}) = \frac{1}{288}((\mathcal{C}^2)^\rho_\sigma\partial^\gamma\otimes x_\rho\mathcal{C}^\sigma_\gamma - (\mathcal{C}^2)^\rho_\tau\partial^\gamma \otimes x_\rho\mathcal{C}^\tau_\alpha)=0

Thus, the contribution in the third order is zero and all up to third order we get that the right polarized twist has exponent

12𝒞 ν ρL ρ ν112𝒞 λ σL ρ λ𝒞 σ ρ124(𝒞 2) ρ νL ρ ν -\frac{1}{2}\mathcal{C}^\rho_\nu\otimes L^\nu_\rho -\frac{1}{12}\mathcal{C}^\sigma_\lambda\otimes L^\lambda_\rho \mathcal{C}^\rho_\sigma -\frac{1}{24}(\mathcal{C}^2)^\nu_\rho\otimes L_\rho^\nu

and for the left polarized twist L 1=exp(x ρ ρ)exp(y^ α α)\mathcal{F}^{-1}_L = exp(-x_\rho\otimes\partial^\rho)exp(\hat{y}_\alpha\otimes\partial^\alpha) the exponent is

+12L ρ ν𝒞 ν ρ112L ρ λ𝒞 σ ρ𝒞 λ σ124L ρ ν(𝒞 2) ρ ν +\frac{1}{2}L^\nu_\rho\otimes\mathcal{C}^\rho_\nu -\frac{1}{12}L^\lambda_\rho \mathcal{C}^\rho_\sigma\otimes\mathcal{C}^\sigma_\lambda -\frac{1}{24}L_\rho^\nu\otimes(\mathcal{C}^2)^\nu_\rho

and if we add the two exponent and divide by 2 we get different coefficients in the 4 second order terms than above.

Last revised on October 24, 2014 at 14:24:28. See the history of this page for a list of all contributions to it.